$$\sum \frac{(n-\ln(n))^n}{2^n \cdot n^n}$$
$$u_{n}= \frac{(n-\ln(n))^n}{2^n \cdot n^n}$$
$$\Rightarrow (u_n)^{1/n} = \frac{n- \ln(n)}{2\cdot n } = \frac12\cdot (1 - \frac{\ln(n)}{n}) \to \frac12 \cdot(1-0) = \frac12 \;\; as \; n \to \infty$$
Hence convergent by root test.
Is this proof fine?
It is correct. You may also say that $$0<\frac{(n-\ln(n))^n}{2^n \cdot n^n}=\frac{(1-\frac{\ln(n)}{n})^n}{2^n}\leq \frac{1}{2^n}$$ and use the comparison test.